/*
 * @lc app=leetcode.cn id=34 lang=java
 *
 * [34] 在排序数组中查找元素的第一个和最后一个位置
 *
 * https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/description/
 *
 * algorithms
 * Medium (38.25%)
 * Likes:    272
 * Dislikes: 0
 * Total Accepted:    50.4K
 * Total Submissions: 131.6K
 * Testcase Example:  '[5,7,7,8,8,10]\n8'
 *
 * 给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
 *
 * 你的算法时间复杂度必须是 O(log n) 级别。
 *
 * 如果数组中不存在目标值，返回 [-1, -1]。
 *
 * 示例 1:
 *
 * 输入: nums = [5,7,7,8,8,10], target = 8
 * 输出: [3,4]
 *
 * 示例 2:
 *
 * 输入: nums = [5,7,7,8,8,10], target = 6
 * 输出: [-1,-1]
 *
 */

// @lc code=start
class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] dp = new int[2];
        dp[0] = -1; dp[1] = -1;
        int left = 0, right = nums.length - 1, mid = (left + right) >> 1;
        while(left <= right) {
            mid = (left + right) >> 1;
            if(nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                if(mid - 1 < 0) {
                    dp[0] = 0;
                    break;
                } else{
                    if(nums[mid-1] < target) {
                        dp[0] = mid;
                        break;
                    } else {
                        right = mid - 1;
                    }
                }
            }
        }
        if(dp[0] == -1) {
            return dp;
        }
        left = 0; right = nums.length - 1;
        while(left <= right) {
            mid = (left + right) >> 1;

            if(nums[mid] < target) {
                left = mid + 1;

            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                if(mid + 1 > nums.length - 1) {
                    dp[1] = nums.length - 1;
                    break;
                } else{
                    if(nums[mid+1] > target) {
                        dp[1] = mid;
                        break;
                    } else {
                        left = mid + 1;
                    }
                }
            }
        }
        return dp;
    }
}
// @lc code=end

